119_pdfsam_math 54 differential equation solutions odd

119_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Exercises 3.2 problems.) Thus we use the general idea in getting a differential equation describing this process: rate of change = input rate − output rate. (3.4) The “input rate” is the rate of mass coming from Hh’s decay, which is opposite to the rate of decay of Hh (Hh looses the mass but It gains it), i.e., input rate = − dHh = 2e−2t , dt (3.5) where we have used (3.3). The “output rate” is the rate with which It decays, which (again, according to the general law of a radioactive decay) is proportional to its current mass. Since the decay constant for It is k = kIt = −1/yr, output rate = kIt It(t) = −It(t). Therefore, combining (3.4)–(3.6) we get the equation for It, that is, dIt(t) = 2e−2t − It(t) dt ⇒ dIt(t) + It(t) = 2e−2t . dt (1)dt = et . (3.6) This is a linear equation with P (t) ≡ 1 and an integrating factor µ(t) = exp Multiplying the equation by µ(t) yields d [et It(t)] = 2e−t dt ⇒ et It(t) = −2e−t + C ⇒ It(t) = −2e−2t + Ce−t . Initially, there were no It, which means that It(0) = 0. With this initial condition we find that 0 = It(0) = −2e−2(0) + Ce−(0) = −2 + C and the mass of It remaining after t years is It(t) = −2e−2t + 2e−t = 2 e−t − e−2t . (3.7) ⇒ C = 2, The element Bu only gains its mass from It, and the rate with which it does this is opposite to the rate with which It looses its mass. Hence (3.6) yields dBu(t) = It(t) = 2 e−t − e−2t . dt 115 ...
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